hdu 2639 （0-1背包第k优解）Bone Collector II

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本文探讨了骨收集器II问题，旨在找到骨头集合中第K大的骨头价值。通过实例输入和输出，解释了解决方法，并提供了解题代码。重点在于算法设计，包括使用结构体和队列来优化解决方案。

Bone Collector II

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 821 Accepted Submission(s): 390

Problem Description

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link:

Here is the link: http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.

If the total number of different values is less than K,just ouput 0.

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the K-th maximum of the total value (this number will be less than 2 ³¹).

Sample Input

Sample Output

Author

teddy

Source

百万秦关终属楚

Recommend

teddy

尝试用结构体来储存，到最后再排序，但是感觉会有的情况是不存在的，所以还是用队列标记再合并：

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
int val[1005],v[1005];
int f[1005][35],A[35],B[35];
int N,V,K;
void dp()
{
    int i,j,k,a,b,c;
    memset(f,0,sizeof(f));
    for(i=1;i<=N;i++)
    for(j=V;j>=v[i];j--)
    {
        for(k=1;k<=K;k++)
        {
           A[k]=f[j-v[i]][k]+val[i];
           B[k]=f[j][k];
        }
        A[k]=-1;B[k]=-1;
        a=b=c=1;
        while(c<=K&&(A[a]!=-1||B[b]!=-1))
        {
            if(A[a]>B[b])
            f[j][c]=A[a],a++;
            else
            f[j][c]=B[b],b++;
            if(f[j][c]!=f[j][c-1])
            c++;
        }
    }
    cout<<f[V][K]<<endl;
}
int main()
{
    int t,i;
    cin>>t;
    while(t--)
    {
        cin>>N>>V>>K;
        for(i=1;i<=N;i++)
        cin>>val[i];
        for(i=1;i<=N;i++)
        cin>>v[i];
        dp();
    }
    return 0;
}